Time needed for velocity to become v<vf: t = (1/gy) ln (Vf/v) + (Vf/v) - 1

[s4orce]

1. When a particle falls though the air, its initial acceleration a=g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If the variation of the acceleration can be expressed as a=(g/v^2f)(v^2f-V62), determine the time needed for the velocity to become v<vf. Initially the particle falls from rest.

Need help...

 

[berkeman]

determine the time needed for the velocity to become v<vf. Initially the particle falls from rest.

Do you mean find the time for the velocity to become v=vf? v will be < vf all the way up until the particle reaches terminal velocity.

And to solve the problem, you will need to do the explicit integrations. The only reason that the simple kinematic equations of motion do not have integrals in them is becuase the acceleration is constant with time. When it is changing, you need to use the integral forms of those equations. Can you show us how you would set those up?

 

[s4orce]

It is v< (less than) vf.

What I have thus far is

a=g(Vf^2-V2^2/Vf^2)

a=g(1-(V2^2/Vf^2)

Integral 0 to V (1/1-(V^2/Vf^2) dv=Integral 0 to T gdt

1/2 ln V+Vf/V-Vs=gt

t=1/gy ln V+Vt/V-Vt

dv/dt=g(1-V^2/Vf^2)

1/2a ln v+a/v-a

1/2 ln V/Vf+1 / V/ Vx-1 Integral 0 to V gt